Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <cstdio>
using namespace std;

const long long B=1000000000000000003LL;
const long long P=1000000000000000031LL;

long long pow_binary_mod(long long x,long long y,long long mod){
	long long z=1;
	for(;y;y>>=1){
		if((y&1)!=0)z=(__int128_t)z*x%mod;
		x=(__int128_t)x*x%mod;
	}
	return z;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N;
	string s;
	cin>>N>>s;
	long long Brev=pow_binary_mod(B,P-2,P);
	vector<long long> o(N+1),p(N+1);p[0]=1;
	for(int i=1;i<=N;i++){
		p[i]=p[i-1];o[i]=o[i-1];
		if(s[i-1]=='<'){
			p[i]=(__int128_t)p[i]*Brev%P;
		}else if(s[i-1]=='>'){
			p[i]=(__int128_t)p[i]*B%P;
		}else if(s[i-1]=='+'){
			o[i]=(o[i]+p[i])%P;
		}else if(s[i-1]=='-'){
			o[i]=(o[i]-p[i]+P)%P;
		}
	}
	map<long long,int>m;
	long long r=0;
	for(int i=N;i>=1;i--){
		m[o[i]]++;
		r+=m[((__int128_t)o[N]*p[i-1]+o[i-1])%P];
	}
	printf("%lld\n",r);
}